Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $t = \dfrac{r^2 - 4}{2r^2 - 18r + 28} \times \dfrac{5r - 35}{-2r + 2} $
Solution: First factor out any common factors. $t = \dfrac{r^2 - 4}{2(r^2 - 9r + 14)} \times \dfrac{5(r - 7)}{-2(r - 1)} $ Then factor the quadratic expressions. $t = \dfrac {(r - 2)(r + 2)} {2(r - 2)(r - 7)} \times \dfrac {5(r - 7)} {-2(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac { (r - 2)(r + 2) \times 5(r - 7)} { 2(r - 2)(r - 7) \times -2(r - 1)} $ $t = \dfrac {5(r - 2)(r + 2)(r - 7)} {-4(r - 2)(r - 7)(r - 1)} $ Notice that $(r - 2)$ and $(r - 7)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {5\cancel{(r - 2)}(r + 2)(r - 7)} {-4\cancel{(r - 2)}(r - 7)(r - 1)} $ We are dividing by $r - 2$ , so $r - 2 \neq 0$ Therefore, $r \neq 2$ $t = \dfrac {5\cancel{(r - 2)}(r + 2)\cancel{(r - 7)}} {-4\cancel{(r - 2)}\cancel{(r - 7)}(r - 1)} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $t = \dfrac {5(r + 2)} {-4(r - 1)} $ $ t = \dfrac{-5(r + 2)}{4(r - 1)}; r \neq 2; r \neq 7 $